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An optimal dividends problem with a terminal value for spectrally negative Lévy processes with a completely monotone jump density R.L. Loeffen ∗ Radon Institute for Computational and Applied Mathematics, Austrian Academy of Sciences November 25, 2008 Abstract We consider a modified version of the classical optimal dividends problem of de Finetti in which the objective function is altered by adding in an extra term which takes account of the ruin time of the risk process, the latter being modeled by a spectrally negative Lévy process. We show that, with the exception of a small class, a barrier strategy forms an optimal strategy under the condition that the Lévy measure has a completely monotone density. As a prerequisite for the proof we show that under the aforementioned condition on the Lévy measure, the q-scale function of the spectrally negative Lévy process has a derivative which is strictly log-convex. AMS 2000 subject classifications. Primary 60J99; secondary 93E20, 60G51. Keywords. Lévy process, stochastic control, dividend problem, scale function, complete monotonicity. 1 Introduction In this paper we consider the classical de Finetti’s optimal dividends problem but with an extra component regarding the ruin time added to the objective function. Within this problem we assume that the underlying dynamics of the risk process is described by a spectrally negative Lévy process which is now widely accepted and used as a replacement for the classical Cramér-Lundberg process (cf. [1, 3, 8, 9, 11, 14, 16, 19, 20, 23]). Recall that a Cramér-Lundberg risk ∗ Postal address: Altenbergerstrasse email: [email protected] 1 69, A-4040 Linz, Austria. process {Xt : t ≥ 0} corresponds to Xt = x + ct − Nt X Ci , i=1 where x > 0 denotes the initial surplus, the claims C1 , C2 , . . . are i.i.d. positive random variables with expected value µ, c > 0 represents the premium rate and N = {Nt : t ≥ 0} is an independent Poisson process with arrival rate λ. Traditionally it is assumed in the Cramér-Lundberg model that the net profit condition c > λµ holds, or equivalently that X drifts to infinity. In this paper X will be a general spectrally negative Lévy process and the condition that X drifts to infinity will not be assumed. We will now state the control problem considered in this paper. As mentioned before, X = {Xt : t ≥ 0} is a spectrally negative Lévy process which is defined on a filtered probability space (Ω, F, F = {Ft : t ≥ 0}, P) satisfying the usual conditions. Within the definition of a spectrally negative Lévy process it is implicitly assumed that X does not have monotone paths. We denote by {Px , x ∈ R} the family of probability measures corresponding to a translation of X such that X0 = x, where we write P = P0 . Further Ex denotes the expectation with respect to Px with E being used in the obvious way. The Lévy triplet of X is given by (γ, σ, ν), where γ ∈ R, σ ≥ 0 and ν is a measure on (0, ∞) satisfying Z 1 ∧ x2 ν(dx) < ∞. (0,∞) Note that even though X only has negative jumps, for convenience we choose the Lévy measure to have only mass on the positive instead of the negative half line. The Laplace exponent of X is given by Z 1 ψ(θ) = log E eθX1 = γθ + σ 2 θ2 − 1 − e−θx − θx1{0<x<1} ν(dx) 2 (0,∞) and is well defined for θ ≥ 0. Note that the Cramér-Lundberg process corresponds to R the case that σ = 0, ν(dx) = λF (dx) where F is the law of C1 and γ = c − (0,1) xν(dx). The process X will represent the risk/surplus process of an insurance company before dividends are deducted. We denote a dividend or control strategy by π, where π = {Lπt : t ≥ 0} is a non-decreasing, left-continuous F-adapted process which starts at zero. The random variable Lπt will represent the cumulative dividends the company has paid out until time t under the control π. We define the controlled (net) risk process U π = {Utπ : t ≥ 0} by Utπ = Xt − Lπt . Let σ π = inf{t > 0 : Utπ < 0} be the ruin time and define the value function of a dividend strategy π by "Z π # σ e−qt dLπt + Se−qσ vπ (x) = Ex π , 0 where q > 0 is the discount rate and S ∈ R is the terminal value. By definition it follows that vπ (x) = S for x < 0. A strategy π is called admissible if ruin 2 does not occur due to a lump sum dividend payment, i.e. Lπt+ − Lπt ≤ Utπ ∨ 0 for t ≤ σ π . Let Π be the set of all admissible dividend policies. The control problem consists of finding the optimal value function v∗ given by v∗ (x) = sup vπ (x) π∈Π and an optimal strategy π∗ ∈ Π such that vπ∗ (x) = v∗ (x) for all x ≥ 0. When S = 0 the above optimal control problem transforms, albeit within the more general framework of a spectrally negative Lévy risk process, to the original optimal dividends problem introduced firstly in a discrete time setting by de Finetti [7] and later studied in, amongst others, [3, 4, 12, 19, 20]. The general case when S ∈ R we consider here is not new. Thonhauser and Albrecher [26] have studied in the Cramér-Lundberg setting the case S < 0. In that case the extra term added to the value function penalizes early ruin and so this model can be used if, besides the value of the dividend payments, one also wants to take into consideration the lifetime of the risk process. The parameter S can then be used to find the desired ’balance’ between optimizing the value of the dividends and maximizing the ruin time. When S > 0, the model can be used if the company, when it becomes bankrupt, has a salvage value equaling S which is distributed to the same beneficiaries as the dividends are, see also the discussion in Radner and Shepp [22, Section 3]. In a Brownian motion/diffusion setting this control problem has been studied in [5, 24]. We will now introduce two types of dividend strategies and state our main theorem. We denote by πa = {Lat : t ≥ 0} the barrier strategy at level a ≥ 0 with corresponding value function va and ruin time σ a . This strategy is defined by La0 = 0 and Lat = sup Xs − a ∨ 0 for t > 0. 0≤s<t Note that πa ∈ Π. So if dividends are paid out according to a barrier strategy with the barrier placed at a, then the corresponding controlled risk process will be a spectrally negative Lévy process reflected in a. We further introduce the take-the-money-and-run strategy πrun = {Lrun : t t ≥ 0} which is the strategy where directly all of the surplus of the company is paid out and immediately thereafter ruin is forced (note that ruin is defined as the state when the controlled risk process is strictly below zero). The value of this strategy is vrun (x) = x + S for x ≥ 0. In case X is not a CramérLundberg risk process, this strategy is the same as the barrier strategy with the barrier placed at zero (i.e. almost surely, L0t = Lrun for all t ≥ 0). But if X t is a Cramér-Lundberg risk process, then the barrier strategy at zero does not imply immediate ruin; ruin occurs only after the first jump/claim which takes an exponentially distributed with parameter ν(0, ∞) amount of time. Therefore the value of the latter strategy might be different than the value of the take-themoney-and-run strategy. In particular for large terminal values, vrun might be 3 bigger than v0 since it can be beneficial to become ruined as soon as possible. Note that in the Cramér-Lundberg case, ruin can be forced in an admissible way by paying out dividends at a rate which is larger than the premium rate immediately after taking out all the surplus. Recall that an infinitely differentiable function f : (0, ∞) → [0, ∞) is completely monotone if its derivatives alternate in sign, i.e. (−1)n f (n) (x) ≥ 0 for all n = 0, 1, 2, . . . for all x > 0. The main theorem of this paper reads now as follows. Theorem 1. Suppose the Lévy measure of the spectrally negative Lévy process X with Lévy triplet (γ, σ, ν), has a completely monotone density. Let c = γ + R1 xν(dx). Then the following holds. 0 (i) If σ > 0, or ν(0, ∞) = ∞, or ν(0, ∞) < ∞ and S ≤ c/q, then an optimal strategy for the control problem is formed by a barrier strategy. (ii) If σ = 0 and ν(0, ∞) < ∞ and S > c/q, then the take-the-money-and-run strategy is an optimal strategy for the control problem. Note that the parameter c is strictly positive since we assumed that the paths of X are not monotone decreasing. For X being equal to a Brownian motion with drift, this control problem has been solved in [5, 24]. In the case when X is a Cramér-Lundberg process with exponentially distributed claims, the control problem was solved by Gerber [12] for S = 0 and by Thonhauser and Albrecher [26] for S < 0 (note that Thonhauser and Albrecher [26, Lemma 10] distinguish between the cases of the optimal strategy being a barrier strategy where the barrier is placed at zero and where the barrier is placed at a strictly positive level, whereas we in Theorem 1 distinguish between the cases where a barrier strategy is optimal and where the take-the-money-and-run strategy is optimal). Note that both cases are examples for which the Lévy measure has a completely monotone density and thus are contained in Theorem 1. Some other examples of spectrally negative Lévy processes which have a Lévy measure with a completely monotone density can be found in [20]. Building on the work of Avram et al. [3], Loeffen [20] proved Theorem 1 for S = 0. In particular, it was shown that optimality of the barrier strategy depends on the shape of the so-called scale function of a spectrally negative Lévy process. To be more specific, the q-scale function of X, W (q) : R → [0, ∞) where q ≥ 0, is the unique function such that W (q) (x) = 0 for x < 0 and on [0, ∞) is a strictly increasing and continuous function characterized by its Laplace transform which is given by Z ∞ 1 e−θx W (q) (x)dx = for θ > Φ(q), ψ(θ) − q 0 where Φ(q) = sup{θ ≥ 0 : ψ(θ) = q} is the right-inverse of ψ. Loeffen [20] showed that when W (q) is sufficiently smooth and W (q)0 is increasing on (a∗ , ∞) where a∗ is the largest point where W (q)0 attains its global minimum, then the barrier strategy at a∗ is optimal for the control problem (in the S = 0 case). 4 Here W (q) being sufficiently smooth means that W (q) is once/twice continuously differentiable when X is of bounded/unbounded variation. It was then shown in [20] that when X has a Lévy measure which has a completely monotone density, these conditions on the scale function are satisfied and in particular that W (q)0 is strictly convex on (0, ∞). Shortly thereafter, Kyprianou et al. [19] showed that W (q)0 is strictly convex on (a∗ , ∞) (but not necessarily on (0, ∞), see [19, Section 3]) under the weaker condition that the Lévy measure has a density which is log-convex. Though the scale function is in that case not necessarily sufficiently smooth, Kyprianou et al. [19] were able to circumvent this problem and proved that the barrier strategy at a∗ is still optimal when the Lévy measure has a log-convex density. Note that without a condition on the Lévy measure the barrier strategy is not optimal in general. Indeed Azcue and Muler [4] have given an example for which no barrier strategy is optimal. The proof of Theorem 1 in the case when S 6= 0, relies on the assumption that W (q)0 is strictly log-convex on (0, ∞). Though in [20] it was only shown under the complete monotonicity assumption on the Lévy measure, that W (q)0 is strictly convex on (0, ∞), we will show in Section 2 that the stronger property of strict log-convexity actually holds in that case. Then in Section 3 the proof of Theorem 1 will be given. 2 Scale functions Associated to the functions {W (q) : q ≥ 0} mentioned in the previous section are the functions Z (q) : R → [1, ∞) defined by Z x Z (q) (x) = 1 + q W (q) (y)dy 0 for q ≥ 0. Together, the functions W (q) and Z (q) are collectively known as scale functions and predominantly appear in almost all fluctuation identities for spectrally negative Lévy processes. As an example we mention the one sided exit below problem for which − q Ex e−qτ0 1(τ − <∞) = Z (q) (x) − W (q) (x), (1) 0 Φ(q) where τ0− = inf{t > 0 : Xt < 0}. We will now recall some properties of scale functions which we will need later on. When the Lévy process drifts to infinity or equivalently ψ 0 (0+) > 0, the 0scale function W (0) (which will be denoted from now on by W ) is bounded and has a limit limx→∞ W (x) = 1/ψ 0 (0+). Further for q ≥ 0 there is the following relation between scale functions W (q) (x) = eΦ(q)x WΦ(q) (x), (2) where WΦ(q) is the (0-)scale function of X under the measure PΦ(q) defined by dPΦ(q) = eΦ(q)Xt −qt . dP Ft 5 The process X under the measure PΦ(q) is still a spectrally negative Lévy process and its Laplace exponent is given by ψΦ(q) (θ) = ψ(Φ(q) + θ) − ψ(Φ(q)). When 0 q > 0 it is known that ψΦ(q) (0+) = ψ 0 (Φ(q)) > 0. From [15, (8.18)] and the fact that W is strictly positive on (0, ∞), it follows that we can write for x, a > 0 Z x log(W (x)) = log(W (a)) + g(t)dt, a where g is a decreasing function and hence log(W (x)) is concave on (0, ∞) (see e.g. [27, Theorem 1.13]). Recall here that a strictly positive function f is said to be log-concave (log-convex) whenever log(f ) is concave (convex). From (2) it now follows that for q ≥ 0, log(W (q) (x)) is concave on (0, ∞) and thus W (q) is log-concave on (0, ∞) for all q ≥ 0. The initial value of the scale function W (q) (0) is equal to 1/c, where c is as in Theorem 1. Note that if X is of unbounded variation, then c = ∞ and thus W (q) (0) = 0. The initial value of the derivative of the scale function is given by (see e.g. [18]) 2 when σ > 0 2/σ (q)0 (q)0 W (0) := lim W (x) = (ν(0, ∞) + q)/c2 when σ = 0 and ν(0, ∞) < ∞ x↓0 ∞ otherwise. Despite the fact that the scale function is in general only implicitly known through its Laplace transform, there are plenty examples of spectrally negative Lévy processes for which there exists closed-form expressions for their scale functions, although most of these examples only deal with the q = 0 scale function. In case no explicit formula for the scale function exists, one can use numerical methods as described in [25] to invert the Laplace transform of the scale function. We refer to the papers [13, 17, 19] for an updated account on explicit examples of scale functions and their properties. In theRsequel for a ∈ R, aR function f and a Borel measure µ, we will use the ∞ ∞ notation a f (x)µ(dx) and a+ f (x)µ(dx) to mean integration over the interval [a, ∞) in the first R ∞ case and integration over the R ∞ interval (a, ∞) in the second case. In particular, a f (x)µ(dx) = f (a)µ{a}+ a+ f (x)µ(dx). We recall Bernstein’s theorem (cf. [10, Chapter XIII.4]) which says that a real-valued function f is completely R ∞ monotone if and only if there exists a Borel measure µ such that f (x) = 0 e−xt µ(dt), x > 0. We now strengthen the conclusion of Theorem 3 in [20]. First we need the following proposition. Proposition 2. Suppose q > 0. Then 0 lim inf eΦ(q)x WΦ(q) (x) = 0. x→∞ Proof. Taking derivatives on both sides in (1) and using (2), we get − d q Φ(q)x 0 Ex e−qτ0 1(τ − <∞) = − e WΦ(q) (x). 0 dx Φ(q) 6 Suppose now that the conclusion of the proposition does not hold. Then 0 eΦ(q)x WΦ(q) (x) will eventually be bounded from below by a strictly positive constant. It follows then that − lim Ex e−qτ0 1(τ − <∞) = −∞, x→∞ 0 which contradicts the positivity of the expectation. Theorem 3. Suppose the Lévy measure ν has a completely monotone density and q > 0. Then the q-scale function can be written as W (q) (x) = eΦ(q)x − f (x), ψ 0 (Φ(q)) x > 0, where f is a completely monotone function. Proof. It was shown in [20] that if the Lévy measure ν has a completely monotone density, then WΦ(q) is a Bernstein function and therefore admits the representation Z ∞ WΦ(q) (x) = a + bx + (1 − e−xt )ξ(dt) x > 0, (3) 0+ R∞ where a, b ≥ 0 and ξ is a measure on (0, ∞) satisfying 0+ (t ∧ 1)ξ(dt) < ∞. Since q > 0, WΦ(q) will be bounded and therefore b = 0 and by using Fatou’s lemma Z ∞ Z ∞ −xt ξ(0, ∞) = lim (1 − e )ξ(dt) ≤ lim (1 − e−xt )ξ(dt) 0+ x→∞ x→∞ 0+ = lim WΦ(q) (x) − a < ∞. x→∞ We now deduce from Proposition 2, (3) and Fatou’s lemma Z ∞ 0 0 = lim inf eΦ(q)x WΦ(q) (x) = lim inf e−x(t−Φ(q)) tξ(dt) x→∞ x→∞ 0+ Z ∞ −x(t−Φ(q)) ≥ lim inf e tξ(dt) ≥ Φ(q)ξ(0, Φ(q)]. 0+ x→∞ It follows that ξ(0, Φ(q)] = 0 and using (2) and (3), we can write Z ∞ W (q) (x) = eΦ(q)x (a + ξ(Φ(q), ∞)) − e−x(t−Φ(q)) ξ(dt) Φ(q)x =e Φ(q)+ ∞ −xt (4) Z (a + ξ(Φ(q), ∞)) − e ξ(dt + Φ(q)). 0+ Now the conclusion of the theorem follows by Bernstein’s theorem and the fact that a + ξ(Φ(q), ∞) = limx→∞ WΦ(q) (x) = 1/ψ 0 (Φ(q)). 7 Denote by W (q,n) (x) the n-th derivative of W (q) (x) for x > 0 and n = 0, 1, 2, . . .. Corollary 4. Suppose the Lévy measure ν has a completely monotone density, q > 0 and n is an odd integer. Then log W (q,n) (x) has a strictly positive second derivative for all x > 0. Consequently, the function W (q,n) is strictly log-convex on (0, ∞). Proof. Suppose that the Lévy measure has a completely monotone density, Φ(q)x q > 0 and n is an odd integer. Let f (x) = ψe0 (Φ(q)) −W (q) (x) and g(x) = −f 0 (x). By Theorem 3, f and g are completely monotone functions and W (q,n) (x) = Φ(q)n Φ(q)x e + g (n−1) (x), ψ 0 (Φ(q)) (5) where g (n−1) is the (n−1)-th derivative of g. Since n is odd, g (n−1) is completely monotone and hence positive (in the weak sense). Further, as q > 0, Φ(q) > 0 and therefore W (q,n) (x) > 0 for x > 0. This means that the following is well defined for x > 0, 2 h i00 hn (x) = W (q,n) (x) log W (q,n) (x) 2 = W (q,n) (x)W (q,n+2) (x) − W (q,n+1) (x) . We need to prove that hn (x) > 0 for all x > 0. Using (5) we get 2 (n−1) (n+1) (n) hn (x) = g (x)g (x) − g (x) o (6) Φ(q)n Φ(q)x n 2 (n−1) (n+1) (n) e Φ(q) g (x) + g (x) − 2Φ(q)g (x) . + 0 ψ (Φ(q)) By Bernstein’s theorem and Hölder’s inequality we have for any completely monotone function v that for some Borel measure µ Z ∞ 2 Z ∞ Z ∞ 00 0 2 −xt 2 −xt −xt v(x)v (x)−(v (x)) = e µ(dt) t e µ(dt)− te µ(dt) ≥ 0 0 0 (n−1) 0 and therefore v is log-convex. Since g is completely monotone, it is logconvex and therefore the expression between the square brackets in (6) is positive. Further, the complete monotonicity of g (n−1) implies that each of the terms between the curly brackets in (6) is positive and hence hn (x) ≥ 0. As Φ(q) > 0, it suffices to prove that one of the terms between the curly brackets, say g (n+1) (x), is strictly positive. We do this by contradiction. Suppose g (n+1) (x) = 0. Then it is easily seen from Bernstein’s theorem that the function f has to be equal to a constant. In that case (4) implies that f ≡ 0. But this means that for λ > Φ(q) Z ∞ Z ∞ −(λ−Φ(q))x 1 e 1 −λx (q) = e W (x)dx = dx = . 0 (Φ(q)) 0 (Φ(q)) ψ(λ) − q ψ (λ − Φ(q))ψ 0 0 8 Thus ψ(λ) is the Laplace exponent of a subordinator (consisting of just a single drift term). But subordinators were excluded from the definition of a spectrally negative Lévy process, which gives us the desired contradiction. 3 Proof of main theorem In this section the proof of Theorem 1 will be given with the aid of a series of lemmas. The approach is similar to [3] and [20], namely calculating the value of a barrier strategy where the barrier is arbitrary, then choosing the ‘optimal’ barrier and finally putting this particular barrier strategy (or the takethe-money-and-run strategy) through a verification lemma. First we recall what we mean by the term sufficiently smooth. A function f : R → R which vanishes on (−∞, 0) and which is right-continuous at zero, is called sufficiently smooth at a point x > 0 if f is continuously differentiable at x when X is of bounded variation and is twice continuously differentiable at x when X is of unbounded variation. A function is then called sufficiently smooth if it is sufficiently smooth at all x > 0. Note that we implicitly assume that a sufficiently smooth function is right-continuous at zero and that it equals zero on (−∞, 0). We let Γ be the operator acting on sufficiently smooth functions f , defined by Z ∞ σ 2 00 0 f (x) + [f (x − y) − f (x) + f 0 (x)y1{0<y<1} ]ν(dy). Γf (x) = γf (x) + 2 0+ Lemma 5 (Verification lemma). Suppose π̂ is an admissible dividend strategy such that (vπ̂ − S) is sufficiently smooth, vπ̂ (0) ≥ S and for all x > 0 max{Γvπ̂ (x) − qvπ̂ (x), 1 − vπ̂0 (x)} ≤ 0. (7) Then vπ̂ (x) = v∗ (x) for all x ≥ 0 and hence π̂ is an optimal strategy. Proof. By definition of v∗ , it follows that vπ̂ (x) ≤ v∗ (x) for all x ≥ 0. We write w := vπ̂ and show that w(x) ≥ vπ (x) for all π ∈ Π for all x ≥ 0. First we suppose x > 0. We define for π ∈ Π the stopping time σ0π by σ0π = inf{t > 0 : Utπ ≤ 0} and denote by Π0 the following set of admissible dividend strategies Π0 = {π ∈ Π : σ0π = σ π Px -a.s. for all x > 0}. Note that when X is of unbounded variation, Π0 = Π, but that Π0 is a strictly smaller set than Π when X is of bounded variation. We claim that any π ∈ Π can be approximated by dividend strategies from Π0 in the sense that for all > 0 there exists π ∈ Π0 such that vπ (x) ≤ vπ (x) + and therefore it is enough to show that w(x) ≥ vπ (x) for all π ∈ Π0 . Indeed, we can take π to be the strategy where you do not pay out any dividends until the stopping time κ := inf{t > 0 : Lπt ≥ }, then at that time point κ pay out a dividend equal to the size of the overshoot of Lπ over and afterwards follow the same strategy as π until ruin occurs for the latter strategy at which point you force 9 ruin immediately. Note that π ∈ Π0 because if σ0π < κ, then σ0π = σ π since until the first dividend payment is made, the process U π is equal to X and for the spectrally negative Lévy process X, the first entry time in (−∞, 0] is equal almost surely to the first entry time in (−∞, 0), provided X0 > 0. Further if σ0π ≥ κ and κ < ∞, then since Uκπ > Uκπ we have σ0π ≥ σ π and thus σ0π = σ π since σ π = σ π on the event {σ0π ≥ κ, κ < ∞} by construction. eπ , U eπ We now assume without loss of generality that π ∈ Π0 and we let L be the right-continuous modifications of Lπ , U π . Note that since the filtration e π and U e π are adapted processes. Let F was assumed to be right-continuous, L etπ > (Tn )n∈N be the sequence of stopping times defined by Tn = inf{t > 0 : U 1 π π e e n or Ut < n }. Since U is a semi-martingale and w is sufficiently smooth in particular w and its derivatives are bounded on [1/n, n] for each n - we can use the change of variables/Itô’s formula (cf. [21, Theorem II.31 & II.32]) on e π ) together with (7) to deduce e−q(t∧Tn ) w(U t∧Tn e0π ) ≥ w(U Z t∧Tn π e πs + e−q(t∧Tn ) w(U et∧T e−qs dL ) + Mt , n 0+ (8) where {Mt : t ≥ 0} is a zero-mean Px -martingale. The details yielding this inequality are given in the Appendix. Using that w ≥ S which follows from the assumptions w(0) ≥ S and w0 ≥ 1, taking expectations, letting t and n go to infinity and using the monotone convergence theorem we get ! Z σπ π π −qs e π e w(U ) ≥ Ex e dLs + SEx e−qσ . 0 0+ Note that we used here that Tn % σ π Px -a.s. which follows because π ∈ Π0 . Now using the mean value theorem together with the assumption that w0 (·) ≥ 1 on (0, ∞), we get e0π ) = w(x − Lπ0+ ) ≤ w(x) − Lπ0+ w(U and combining with ! Z σπ −qs e π Ex e dLs = Ex 0+ Z σπ ! e −qs dLπs π −Lπ0+ = vπ (x)−SEx e−qσ −Lπ0+ , 0 we deduce w(x) ≥ vπ (x) and hence we proved w(x) ≥ v∗ (x) for all x > 0. To finish the proof, note that v∗ is an increasing function (in the weak sence) since the set of admissible strategies is larger when the initial reserves are higher and hence because w is right-continuous at zero, v∗ (0) ≤ limx↓0 v∗ (x) ≤ limx↓0 w(x) = w(0). Proposition 6. Assume W (q) is continuously differentiable on (0, ∞). The value function of the barrier strategy at level a ≥ 0 is given by (q) (a) SZ (q) (x) + W (q) (x) 1−qSW if x ≤ a (q)0 W (a) va (x) = (q) (a) x − a + SZ (q) (a) + W (q) (a) 1−qSW if x > a. W (q)0 (a) 10 Proof. Clearly the proposition only needs to be proved for 0 ≤ x ≤ a. Let x ∈ [0, a]. By Avram et al. [3, Proposition 1], it follows that "Z a # σ W (q) (x) −qt a Ex e dLt = (q)0 . W (a) 0 Since a σ = inf{t > 0 : Xt − Lat < 0} = inf{t > 0 : sup Xs ∨ a − Xt > a}, 0≤s<t it follows by Avram et al. [2, Theorem 1] that h i a qW (q) (a) Ex e−qσ = Z (q) (x) − W (q) (x) (q)0 . W (a) Define the function ζ : [0, ∞) → R by ζ(x) = 1 − qSW (q) (x) W (q)0 (x) for x > 0 and ζ(0) = limx↓0 ζ(x). We now define the (candidate) optimal barrier level by a∗ (S) = sup {a ≥ 0 : ζ(a) ≥ ζ(x) for all x ≥ 0} . Hence a∗ (S) is the last point where ζ attains its global maximum. Note that a∗ (0) is the point a∗ mentioned in Section 1. In the sequel we will write a∗ instead of a∗ (0). Proposition 7. Suppose W (q) is continuously differentiable on (0, ∞). Then a∗ (S) < ∞. Proof. Define 0 1 + qS Φ(q)−1 W (q) (x) − W (q) (x) qS f (x) = ζ(x) + = . Φ(q) W (q)0 (x) (q) W (x) 1 (q) Since limx→∞ W is continuously (q)0 (x) = Φ(q) (see e.g. [3, Section 3.3]) and W differentiable, it follows that limx→∞ f (x) = 0 and f is continuous. Hence a∗ (S) < ∞ if there exists x ≥ 0 such that f (x) > 0. But by (2) 1+ f (x) = qS Φ(q)x 0 WΦ(q) (x) Φ(q) e 0 W (q) (x) and thus by Proposition 2, there exists x ≥ 0 such that f (x) > 0. 11 Note that when a∗ (S) > 0 and W (q) is twice continuously differentiable, then ζ (a∗ (S)) = 0. Further, it is easily seen that if an optimal strategy is formed by a barrier strategy, then the barrier strategy at a∗ (S) has to be an optimal strategy. 0 Lemma 8. Suppose W (q) is sufficiently smooth and that ζ(a) ≥ ζ(b) for all a, b such that a∗ (S) ≤ a ≤ b. (9) Then the following holds. (i) If ζ(a∗ (S)) ≥ 0, then the barrier strategy at a∗ (S) is an optimal strategy. (ii) If a∗ (S) = 0 and ζ(0) ≤ 0, then the take-the-money-and-run strategy is optimal. Note that Lemma 8 is a generalization of Theorem 2 in [20]. Indeed when S = 0, ζ(a∗ ) = 1/W (q)0 (a∗ ) > 0 and condition (9) transforms into the condition that W (q)0 is increasing on (a∗ , ∞). Proof. We first prove (i) by showing that va∗ (S) satisfies the conditions of the verification lemma. Using (9), all the conditions of the verification lemma can be proved following the same arguments as in the proofs of Lemma 5 and Theorem 2 in [20], with the exception being the condition that va∗ (S) (0) ≥ S. (Note that in deducing the analogue ofequation (4) in [20], one also uses the − fact that e−q(t∧τ0 ∧τa+ ) Z (q) (Xt∧τ − ∧τa+ ) 0 t≥0 is a Px -martingale, cf. [15, p.229].) The missing condition now follows from va∗ (S) (0) = SZ (q) (0) + W (q) (0)ζ(a∗ (S)) ≥ S, where the inequality follows from the assumption that ζ(a∗ (S)) ≥ 0. For case (ii) we prove that vrun satisfies the conditions of the verification lemma. Note that since vrun (x) = x + S for x ≥ 0, the only non-trivial thing to show is that (Γ−q)vrun (x) ≤ 0 for all x > 0. This can be achieved by mimicking the proof of Theorem 2 in [20], which involves proving that lim(Γ − q)(vrun − vx )(y) ≤ 0 y↑x for x > 0. Note that in order to prove the above inequality, one uses that vrun (0) ≥ vx (0) which follows from ζ(x) ≤ 0 and the latter is due to the assumption that ζ(0) ≤ 0 and a∗ (S) = 0 (combined with (9)). Proof of Theorem 1. Since the case S = 0 was proved in Loeffen [20], we assume without loss of generality that S 6= 0. Note that by Theorem 3, W (q) is infinitely differentiable (this was proved for the first time in [6]) and therefore certainly smooth enough. Further note that W (q)00 is strictly negative on (0, a∗ ), 12 strictly positive on (a∗ , ∞) and if a∗ > 0, then W (q)00 (a∗ ) = 0. We will show that ζ is strictly increasing on (0, a∗ (S)) and strictly decreasing on (a∗ (S), ∞), (10) from which it follows that a∗ (S) is the only point where ζ has a local/global maximum and that (9) holds. First note that with g(x) = −qSW (q)0 (x)/W (q)00 (x) for x ∈ (0, ∞)\{a∗ }, the following differential equation holds for ζ ζ 0 (x) = − W (q)00 (x) (ζ(x) − g(x)) , W (q)0 (x) x ∈ (0, ∞)\{a∗ }. From this it follows that for x ∈ (0, a∗ ) ζ 0 (x) > 0(< 0, = 0) iff ζ(x) > g(x)(< g(x), = g(x)), for x ∈ (a∗ , ∞) ζ 0 (x) > 0(< 0, = 0) iff ζ(x) < g(x)(> g(x), = g(x)). Suppose that S > 0. Since h 2 i qS W (q) (x)W (q)00 (x) − W (q)0 (x) − W (q)00 (x) ζ 0 (x) = 2 W (q)0 (x) (11) (12) and the expression between square brackets is negative due to the log-concavity of W (q) , it follows that ζ 0 (x) < 0 on (a∗ , ∞) and therefore a∗ (S) ≤ a∗ . If a∗ = 0, (10) now holds, so we can assume without loss of generality that a∗ > 0. Then limx↑a∗ g(x) = ∞ and (11) imply a∗ (S) 6= a∗ and thus a∗ (S) < a∗ . By the strict log-convexity of W (q)0 (Corollary 4), g is strictly increasing on (0, a∗ ). The foregoing and (11) imply then that either ζ intersects g exactly once on (0, ∞) (at a∗ (S)) and (10) holds or that ζ 0 (x) < 0 for all x > 0 and in that case a∗ (S) = 0. Hence (10) holds when S > 0. Suppose now that S < 0 and a∗ > 0. Then ζ is strictly positive on (0, ∞) by definition and g is strictly negative on (0, a∗ ). Hence a∗ (S) ≥ a∗ . Due to the strict log-convexity of W (q)0 , g is in this case strictly decreasing on (a∗ , ∞) and combined with (11) and the fact that limx↓a∗ g(x) = ∞, this implies that ζ and g intersect each other exactly once, a∗ (S) > a∗ and that (10) holds. This leaves the final case when S < 0 and a∗ = 0. If ζ(0) ≥ g(0), then (11) and g being strictly decreasing on (0, ∞) implies ζ is strictly decreasing on (0, ∞) and hence a∗ (S) = 0. If ζ(0) < g(0), then a∗ (S) > 0 and further (10) holds by the same arguments as before. Suppose now that σ > 0, or ν(0, ∞) = ∞, or ν(0, ∞) < ∞ and S ≤ c/q. Then from the values of W (q) (0) and W (q)0 (0) given in Section 2, it follows that ζ(0) ≥ 0 and hence ζ(a∗ (S)) ≥ 0 by definition of a∗ (S). Thus part (i) of the theorem follows from Lemma 8(i). To prove part (ii), suppose that σ = 0 and ν(0, ∞) < ∞ and S > c/q. This implies S > 0 and ζ(0) < 0. If a∗ = 0 then a∗ (S) = 0 since S > 0. If a∗ > 0 then g(0) > 0 and hence by (11) and (12), ζ 0 (x) < 0 for all x > 0 and therefore a∗ (S) = 0. Part (ii) follows now from Lemma 8(ii). 13 Appendix We give here the details which lead to (8). e π ) gives Applying the change of variables/Itô’s formula to e−q(t∧Tn ) w(U t∧Tn e −q(t∧Tn ) π et∧T e0π ) = w(U ) − w(U n Z t∧Tn e −qs 0+ Z t∧Tn + σ 2 00 e π π e w (Us− ) − qw(Us− ) ds 2 π es− e πs )c ) e−qs w0 (U )d(Xs − (L (13) 0+ + X π esπ ) − w0 (U es− e−qs [∆w(U )∆Xs ], 0<s≤t∧Tn π esπ = U esπ − U es− esπ ) = w(U esπ ) − where we use the following notation: ∆U , ∆w(U P π π π c π e es− ) and (L es ) = L es − w(U 0<u≤s ∆Lu . Note that we have used here the fact that the continuous martingale part of X is its Gaussian part. One can easily verify that X π esπ ) − w0 (U es− e−qs [∆w(U )∆Xs ] = 0<s≤t∧Tn X π π es− es− e−qs [∆w(U + ∆Xs ) − w0 (U )∆Xs ] 0<s≤t∧Tn − X e πs− ) − w(Xs − L e πs− − ∆L e πs )]. (14) e−qs [w(Xs − L 0<s≤t∧Tn e πs ≤ Xs − L e πs− and so by the mean Since by admissibility of Lπ , we have ∆L 0 value theorem and the hypothesis that w ≥ 1, e πs− ) − w(Xs − L e πs− − ∆L e πs ) ≥ ∆L e πs w(Xs − L 14 for 0 < s < t ∧ Tn . (15) Combining (13), (14) and (15) leads to e −q(t∧Tn ) e π ) − w(U e0π ) ≤ w(U t∧Tn Z t∧Tn + Z t∧Tn e −qs 0+ σ 2 00 e π π e w (Us− ) − qw(Us− ) ds 2 π es− e πs )c ) e−qs w0 (U )d(Xs − (L 0+ X + π π es− es− e πs ] e−qs [∆w(U + ∆Xs ) − w0 (U )∆Xs − ∆L 0<s≤t∧Tn Z t∧Tn = e −qs π es− (Γ−q)w(U )ds− 0+ Z t∧Tn π es− e πs )c − e−qs w0 (U )d(L 0+ (Z t∧Tn + X e πs e−qs ∆L 0<s≤t∧Tn ) π es− e−qs w0 (U )d[Xs − γs − 0+ X ∆Xu 1{|∆Xu |≥1} ] 0<u≤s ( X + π π es− es− e−qs [∆w(U + ∆Xs ) − w0 (U )∆Xs 1{|∆Xs |<1} ] 0<s≤t∧Tn Z t∧Tn Z ∞ − 0+ i h ) π π π es− es− es− e−qs w(U − y) − w(U ) + w0 (U )y1{0<y<1} ν(dy)ds . 0+ By the Lévy-Itô decomposition the expression between the first pair of curly brackets is a zero-mean martingale and by the compensation formula (cf. [15, Corollary 4.6]) the expression between the second pair of curly brackets is also a zero-mean martingale. Now using (7), the inequality (8) follows. References [1] H. Albrecher, J.-F. Renaud, and X. 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